Trust gửi ạ 

Đáp án + Giải thích các bước giải:

$a\big)$

$n_{H_2SO_4}=\frac{58,8}{98}=0,6(mol)$

$n_{H_2}=\frac{6,72}{22,4}=0,3(mol)$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2(1)$

$Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O(2)$

Theo PT(1): $n_{H_2SO_4(1)}=0,3(mol);n_{Al}=0,2(mol)$

$\to n_{H_2SO_4(2)}=0,6-0,3=0,3(mo)$

Theo PT(2): $n_{Al_2O_3}=\frac{1}{3}n_{H_2SO_4(2)}=0,1(mol)$

$\to \%m_{Al}=\frac{0,2.27}{0,2.27+0,1.102}.100\%\approx 34,62\%$

$\to \%m_{Al_2O_3}=100-34,62=65,38\%$

$b\big)$

Theo PT: $\sum n_{Al_2(SO_4)_3}=0,5.0,2+0,1=0,2(mol)$

$\to m_1=m_{Al_2(SO_4)_3}=0,2.342=68,4(g)$